Physics

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A battery with an emf of 7.9 V and internal resistance of 1.37 Ω is connected across a load resistor
R.If the current in the circuit is 0.88 A, what is the value of R?
Answer in units of Ω

What power is dissipated in the internal resistance of the battery?
Answer in units of W

How do you set this problem up?

  • Physics -

    I=ℇ/(R+r)
    R= (ℇ-Ir)/I= (7.9-0.88•1.37)/0.88 = …
    P=I²r = …

  • Physics -

    i = 7.9/(1.37+R)
    .88 = 7.9/(1.37+R)
    1.37 + R = 8.98
    so
    R = 7.61 ohms

    P = i^2 r = .88^2(1.37) = 1.06 Watts

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